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Hyperbolic functions help

  1. Aug 10, 2008 #1
    integrate (x^2) / (4+x^2)^(3/2)

    Im not allowed to apply hyperbolic functions to this and have been trying to solve applying to a 90 deg. angle.

    x = 2tan(theta)
    x^2 = 4tan^2(theta)
    dx = 2 sec^2(theta)

    Hopefully you can se where Im going with this (trigonomic substitution)

    Im ready to give up!!!!!!
  2. jcsd
  3. Aug 10, 2008 #2
    Re: stuck!!!!!!!!!

    Hmm that substitution worked for me. I think a similar (or the same) problem appears in Apostol. Anyways try the substitution again. Check your arithmetic and remember that 1+tan^2(x) = sec^2(x). Also don't forget the dx part.
  4. Aug 10, 2008 #3
    Re: stuck!!!!!!!!!

    Im not forgetting about the dx. I have been working on this for 2 hours now, please give me a little more than that. Here is what I have so far that I believe to be right:

    (2)integral tan^2(theta) / sec(theta)
  5. Aug 10, 2008 #4
    Re: stuck!!!!!!!!!

    Remember his said thing
    tan^2 = 1 + sec^2
    Use integration table for finding integral of sec t
  6. Aug 10, 2008 #5
    Re: stuck!!!!!!!!!

    Oh, you've gotten to that part. Well changing everything to sin and cos you get sin^2(theta) / cos(theta). At this point I think it's easier if you write it in terms of tan(theta)*sin(theta) and then try integrating by parts. I chose u to be tan(theta) and dv = sin(theta)d(theta). Either way you'll have to integrate a somewhat obscure but really rather well-known trig expression.

    EDIT, or rootX found an easier way to manipulate that and integrate.
  7. Aug 10, 2008 #6
    Re: stuck!!!!!!!!!

    Wait nevermind, I was wondering why the integration by parts came out so nicely. If you integrated by parts like I did it is equivalent to just convering sin^2(theta) to 1- cos^2(theta) and dividing by cos(theta) you get sec(theta) - cos(theta) so again it comes down to the antiderivative of sec(theta) which you could look up. The derivation requires an insight.
  8. Aug 10, 2008 #7


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    = (sec^2 - 1)/sec = sec - cos :smile:
  9. Aug 10, 2008 #8
    Re: stuck!!!!!!!!!

    okay, so now Ive got:
    (2) integral sec(t) - cos(t)

    Do I go ahead and take the integral at this pont? the integral of sec(t) involves (ln) and I dont beleive that to be correct.
  10. Aug 10, 2008 #9


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    Yes!!! Why not?? :smile:
    i] why?

    ii] try it anyway! :smile:
  11. Aug 10, 2008 #10
    Re: stuck!!!!!!!!!

    Yeah if I remember the derivation correctly you multiply sec(x) by [sec(x) - tan(x)]/[sec(x) - tan(x)] and note that in the resulting expression, the denominator's derivative is the negative of the expression in the numerator. This suggests an integral involving ln.
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